Problem: Simplify the following expression: $y = \dfrac{5x^2+21x+4}{5x + 1}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(5)}{(4)} &=& 20 \\ {a} + {b} &=& &=& {21} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $20$ and add them together. The factors that add up to ${21}$ will be your ${a}$ and ${b}$ When ${a}$ is ${1}$ and ${b}$ is ${20}$ $ \begin{eqnarray} {ab} &=& ({1})({20}) &=& 20 \\ {a} + {b} &=& {1} + {20} &=& 21 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({5}x^2 +{1}x) + ({20}x +{4}) $ Factor out the common factors: $ x(5x + 1) + 4(5x + 1)$ Now factor out $(5x + 1)$ $ (5x + 1)(x + 4)$ The original expression can therefore be written: $ \dfrac{(5x + 1)(x + 4)}{5x + 1}$ We are dividing by $5x + 1$ , so $5x + 1 \neq 0$ Therefore, $x \neq -\frac{1}{5}$ This leaves us with $x + 4; x \neq -\frac{1}{5}$.